INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to a small extent, what one can do with them. Let R be a commutative Noetherian ring with an identity element. An R- module E is injective if HomR( ;E) is an exact functor. The main messages of these notes are − Every R-module M has an injective hull or injective envelope, de- • noted by ER(M), which is an injective module containing M, and has the property that any injective module containing M contains an isomorphic copy of ER(M). A nonzero injective module is indecomposable if it is not the direct • sum of nonzero injective modules. Every injective R-module is a direct sum of indecomposable injective R-modules. Indecomposable injective R-modules are in bijective correspondence • with the prime ideals of R; in fact every indecomposable injective R-module is isomorphic to an injective hull ER(R=p), for some prime ideal p of R. The number of isomorphic copies of ER(R=p) occurring in any direct • sum decomposition of a given injective module into indecomposable injectives is independent of the decomposition. Let (R; m) be a complete local ring and E = ER(R=m) be the injec- • tive hull of the residue field of R. The functor ( )_ = HomR( ;E) has the following properties, known as Matlis duality− : − (1) If M is an R-module which is Noetherian or Artinian, then M __ ∼= M. (2) If M is Noetherian, then M _ is Artinian. (3) If M is Artinian, then M _ is Noetherian. Any unexplained terminology or notation can be found in [1] or [3]. Matlis' theory of injective modules was developed in the paper [4], and may also be found in [3, 18] and [2, 3]. These notes owe a greatx deal of intellectualx debt to Mel Hochster, whose lecture notes are very popular with the organizers of this workshop. How- ever, the organizers claim intellectual property of all errors here. 1. Injective Modules Throughout, R is a commutative ring with an identity element 1 R. All R-modules M are assumed to be unitary, i.e., 1 m = m for all m 2M. 1 · 2 2 INJECTIVE MODULES Definition 1.1. An R-module E is injective if for all R-module homomor- phisms ' : M N and : M E where ' is injective, there exists an R-linear homomorphism−! θ : N −!E such that θ ' = . −! ◦ Exercise 1.2. Show that E is an injective R-module E if and only if HomR( ;E) is an exact functor, i.e., applying HomR( ;E) takes short ex- act sequences− to short exact sequences. − Theorem 1.3 (Baer's Criterion). An R-module E is injective if and only if every R-module homomorphism a E, where a is an ideal, extends to a homomorphism R E. −! −! Proof. One direction is obvious. For the other, if M N are R-modules and ' : M E, we need to show that ' extends to⊆ a homomorphism −! N E. By Zorn's lemma, there is a module N 0 with M N 0 N, which is maximal−! with respect to the property that ' extends to⊆ a homomorphism⊆ ' : N E. If N = N, take an element n N N and consider the ideal 0 0 −! 0 6 2 n 0 n '0 a = N 0 :R n. By hypothesis, the composite homomorphism a N 0 E extends to a homomorphism : R E. Define ' : N +−!Rn −!E by −! 00 0 −! '00(n0 + rn) = '0(n0) + (r). This contradicts the maximality of '0, so we must have N 0 = N. Exercise 1.4. Let R be an integral domain. An R-module M is divisible if rM = M for every nonzero element r R. 2 (1) Prove that an injective R-module is divisible. (2) If R is a principal ideal domain, prove that an R-module is divisible if and only if it is injective. (3) Conclude that Q=Z is an injective Z-module. (4) Prove that any nonzero Abelian group has a nonzero homomorphism to Q=Z. (5) If ( )_ = HomZ( ; Q=Z) and M is any Z-module, prove that the natural− map M − M is injective. −! __ Exercise 1.5. Let R be an A-algebra. (1) Use the adjointness of and Hom to prove that if E is an injective ⊗ A-module, and F is a flat R-module, then HomA(F; E) is an injective R-module. (2) Prove that every R-module can be embedded in an injective R- module. Hint: If M is the R-module, take a free R-module F with a surjection F HomZ(M; Q=Z). Apply ( ) = HomZ( ; Q=Z). −!! − _ − Proposition 1.6. Let M = 0 and N be R-modules, and let θ : M, N be a monomorphism. Then the6 following are equivalent: ! (1) Every nonzero submodule of N has a nonzero intersection with θ(M). (2) Every nonzero element of N has a nonzero multiple in θ(M). (3) If ' θ is injective for a homomorphism ' : N Q, then ' is injective.◦ −! INJECTIVE MODULES 3 Proof. (1) = (2) If n is a nonzero element of N, then the cyclic module Rn has a nonzero) intersection with θ(M). (2) = (3) If not, then ker ' has a nonzero intersection with θ(M), contradicting) the assumption that ' θ is injective. (3) = (1) Let N be a nonzero◦ submodule of N, and consider the ) 0 canonical surjection ' : N N=N 0. Then ' is not injective, hence the composition ' θ : M N=N−! is not injective, i.e., N contains a nonzero ◦ −! 0 0 element of θ(M). Definition 1.7. If θ : M, N satisfies the equivalent conditions of the previous proposition, we say! that N is an essential extension of M. Example 1.8. If R is a domain and Frac(R) is its field of fractions, then R Frac(R) is an essential extension. More generally, if S R is the set ⊆ 1 ⊆ of nonzerodivisors in R, then S− R is an essential extension of R. Example 1.9. Let (R; m) be a local ring and N be an R-module such that every element of N is killed by a power of m. The socle of N is the submodule soc(N) = 0 :N m. Then soc(N) N is an essential extension: if n N is a nonzero element, let t be the smallest⊆ integer such that mtn = 0. Then2 mt 1n soc(N), and mt 1n contains a nonzero multiple of n. − ⊆ − Exercise 1.10. Let I be an index set. Then Mi Ni is essential for all ⊆ i I if and only if i I Mi i I Ni is essential. 2 ⊕ 2 ⊆ ⊕ 2 Example 1.11. Let R = C[[x]] which is a local ring with maximal ideal (x), and take N = Rx=R. Every element of N is killed by a power of the maximal ideal, and soc(N) is the 1-dimensional C-vector space generated by [1=x], i.e., the image of 1=x in N. By Example 1.9, soc(N) N is an essential ⊆ extension. However N soc(N) N N is not an essential extension since the element Q ⊆ Q [1=x]; [1=x2]; [1=x3];::: N Y 2 N does not have a nonzero multiple in N soc(N). (Prove!) Q Proposition 1.12. Let L; M; N be nonzero R-modules. (1) M M is an essential extension. (2) Suppose⊆ L M N. Then L N is an essential extension if and only if both⊆L ⊆M and M N⊆are essential extensions. ⊆ ⊆ (3) Suppose M N and M Ni N with N = iNi. Then M N is ⊆ ⊆ ⊆ [ ⊆ an essential extension if and only if M Ni is an essential extension for every i. ⊆ (4) Suppose M N. Then there exists a module N with M N N, ⊆ 0 ⊆ 0 ⊆ which is maximal with respect to the property that M N 0 is an essential extension. ⊆ Proof. The assertions (1), (2), and (3) elementary. For (4), let = N 0 M N 0 N and N 0 is an essential extension of M : F f j ⊆ ⊆ g 4 INJECTIVE MODULES Then M so is nonempty. If N N N ::: is a chain in 2 F F 10 ⊆ 20 ⊆ 30 ⊆ , then iN is an upper bound. By Zorn's Lemma, the set has F [ i0 2 F F maximal elements. Definition 1.13. The module N 0 in Proposition 1.12 (4) is a maximal es- sential extension of M in N. If M N is essential and N has no proper essential extensions, we say that N is⊆ a maximal essential extension of M. Proposition 1.14. Let M be an R-module. The following conditions are equivalent: (1) M is injective; (2) M is a direct summand of every module containing it; (3) M has no proper essential extensions. Proof. (1) = (2) = (3) is left as an exercise, and we prove the implication (3)) = (2).) Consider an embedding M, E where E is injective. By Zorn's) lemma, there exists a submodule N! E which is maximal with respect to the property that N M = 0. This⊆ implies that M, E=N is an essential extension, and hence\ that it is an isomorphism. But! then E = M + N so E = M N. Since M is a direct summand of an ⊕ injective module, it must be injective. Proposition 1.15. Let M and E be R-modules.